(2)$\implies$(1) always holds: assume that $AB\subseteq P$, and $A$ is not contained in $P$. Then there exists $a\in A$ such that $a\notin P$. But for every $b\in B$, $ab\in AB\subseteq P$, hence by (2) we must have $b\in P$. Thus, $B\subseteq P$. Hence, (2) holds. Note that this holds without the need to assume commutativity of existence of a unity. Condition (2) is called "totally prime" or "completely prime", and is stronger than primality in noncommutative rings.
The problem with (1)$\implies$(2) is just the question of your definition of $(a)$. For commutative rings with unity, $(a)=Ra=\{ra\mid r\in R\}$. However, for commutative rings without unity, since $(a)$ is the smallest ideal that contains $a$, then it is equal to
$$(a) = \{ra + na\mid r\in R,n\in\mathbb{Z}\},$$
where $na$ is defined inductively: $0a=0_R$, $(n+1)a = (na)+a$, and $(-n)a = -(na)$.
In particular, $a\in (a)$ always holds.
(1)$\implies$(2). Assume that $ab\in P$. Then $(ab)\subseteq P$, hence $(a)(b)=(ab)\subseteq P$. Since $P$ satisfies (1), then either $(a)\subseteq P$ or $(b)\subseteq P$. If $(a)\subseteq P$, then $a\in (a)\subseteq P$ so $a\in P$. If $(b)\subseteq P$, then $b\in(b)\subseteq P$, so $b\in P$. This proves (2).
But perhaps you were thinking that $(a)=Ra$ and so were confused about why this argument would work, given that if $R$ has no unity there is no reason to expect that $a\in Ra$? (it might be there anyway even if $R$ has no unity; e.g., in the direct sum of infinitely many copies of $\mathbb{Z}$ there is no unity, but for every $a$ there is an $x$ such that $ax=a$). The argument as is doesn't work, but the implication still holds: just consider $A$ to be the smallest ideal of $R$ that contains $a$, $B$ to be the smallest ideal of $R$ that contains $b$, notice that $ab\in AB$, and then prove, as I do below, that $AB$ is in fact the smallest ideal that contains $ab$.
The condition that $(a)(b)=(ab)$ is established by noting that every element of $(a)(b)$ is a sum of elements of the form $xy$, with $x\in (a)$ and $y\in (b)$. Each of this is of the form
$$(ra+na)(sb+mb) = (rs)ab + (na)sb + ra(mb) + (na)(mb) = (rs + ns + mr)ab + (nm)ab\in (ab).$$
So $(a)(b)\subseteq (ab)$. Since $ab\in (a)(b)$, then $(ab)\subseteq (a)(b)$, which gives the equality.
For not-necessarily-commutative rings, an ideal satisfying condition (2) is known, as I mentioned, as "completely prime" or "totally prime". Completely prime always implies prime, but the converse does not necessarily hold. For an example, take the ring of $n\times n$ matrices with coefficients in a field $F$, $n\gt 1$; the only ideals of this ring are $(\mathbf{0})$ and $R$, so in particular $(\mathbf{0})$ is prime. But it is not completely prime, since for example the matrix $E_{1n}$ ($1$ in the $(1,n)$ coordinate, zeros elsewhere) satisfies $E_{1n}^2 = \mathbf{0}\in (\mathbf{0})$, but $E_{1n}\notin(\mathbf{0})$.
The one in Wikipedia is the standard definition of a simple ring, but from what you describe, Lang's definition amounts to what is more commonly called a simple Artinian ring.
For a simple ring (by which I mean the definition appearing in Wikipedia) the conditions of being right Artinian or left Artinian are equivalent. As a special case of the Artin-Wedderburn theorem, such a ring is just $M_n(D)$ for some division ring $D$.
Perhaps it was Lang's intention to define a simple ring such that all of his simple rings are semisimple too. This is a point of terminological confusion sometimes, because the more common definition allows simple rings which aren't Artinian, hence aren't semisimple.
I don't know if anyone has tried it, but maybe it would make sense to call finite products of simple rings "semisimple" and to give "semisimple Artinian rings" another name, say "Wedderburn ring" or something like that. Isaacs called them Wedderburn rings in his Algebra book, actually, but I don't think it caught anywhere else.
Best Answer
Sure. Let $N$ be any proper right ideal of $R=M_n(F)$ for a field $F$.
Then $N$ satisfies the very first definition: if $A,B$ are ideals of $R$ such that $AB\subseteq N$, then $A$ or $B$ is contained in $N$. For $AB\subseteq N$, one of $A$ or $B$ is zero. (There are only two ideals in $R$!)
There are, however, interesting notions of primeness for one-sided ideals. An especially interesting one is the major one in this paper.